Two lines `L_(1): x=5, (y)/(3-alpha)=(z)/(-2)`
`L_(2):x=alpha, (y)/(-1) =(z)/(2-alpha)` are coplanar. Then, `alpha` can take value (s)

Options

A. 1, 4, 5
B. 1, 2, 5
C. 3, 4, 5
D. 2, 4, 5

Answer

To determine the values of \( \alpha \) for which the lines \( L_1 \) and \( L_2 \) are coplanar, we will use the condition involving the determinant of the direction ratios and the coordinates of points on the lines.

### Step-by-Step Solution:

1. **Identify the equations of the lines:**
The equations of the lines are given as:
\[ L_1: x = 5, \quad \frac{y}{3 - \alpha} = \frac{z}{-2} \]
\[ L_2: x = \alpha, \quad \frac{y}{-1} = \frac{z}{2 - \alpha} \]

2. **Convert the equations into parametric form:**
For line \( L_1 \):
- Let \( z = -2t \) (parameter \( t \))
- Then \( y = (3 - \alpha)t \)
- The parametric form is:
\[ (x, y, z) = (5, (3 - \alpha)t, -2t) \]

For line \( L_2 \):
- Let \( z = (2 - \alpha)s \) (parameter \( s \))
- Then \( y = -s \)
- The parametric form is:
\[ (x, y, z) = (\alpha, -s, (2 - \alpha)s) \]

3. **Set up the determinant condition for coplanarity:**
The lines are coplanar if the following determinant is zero:
\[ \begin{vmatrix} 5 - \alpha & 0 & 0 \\ 3 - \alpha & -1 & 2 - \alpha \\ -2 & 0 & 0 \end{vmatrix} = 0 \]

4. **Calculate the determinant:**
The determinant simplifies to:
\[ (5 - \alpha) \begin{vmatrix} -1 & 2 - \alpha \\ 0 & 0 \end{vmatrix} - 0 + 0 \]
The determinant of the 2x2 matrix is zero, so we focus on the first term:
\[ (5 - \alpha)(-1)(0) - (3 - \alpha)(0) + 0 = 0 \]
This leads to:
\[ (5 - \alpha)(3 - \alpha)(2 - \alpha) = 0 \]

5. **Solve for \( \alpha \):**
The equation \( (5 - \alpha)(3 - \alpha)(2 - \alpha) = 0 \) gives:
\[ \alpha = 5, \quad \alpha = 3, \quad \alpha = 2 \]

Thus, the values of \( \alpha \) for which the lines \( L_1 \) and \( L_2 \) are coplanar are \( \alpha = 2, 3, 5 \).

### Final Answer:
The values of \( \alpha \) are \( 2, 3, 5 \).